Another missing area

[Genady]

[TheVat]

Spoiler

It looks like a rotation of this triangle is really helpful to an easy solution?  Would be 3, if it is indeed an isoceles triangle tipped over.  

 

[Genady]

It is an arbitrary triangle with arbitrary lines in it. No other symmetries assumed.

The three given areas uniquely determine the fourth area.

[Genady]

Like in the previous missing area puzzle (https://www.scienceforums.net/topic/131527-find-the-missing-area/?do=findComment&comment=1239116), it is helpful to switch to triangles:

Spoiler

The area in question is x+y.

 

 

 

[md65536]

Spoiler

7.8?

I made a right triangle that fits the specification, but it got pretty complicated. I haven't figured out how to use the clue!

 

[Genady]

@md65536  Hint:

Spoiler

(4+2)/2 = (3+x+y)/x

 

Edited May 12, 2023 by Genady

[md65536]

I can't imagine ever noticing that, but now I see why it works for other triangles with the same areas. If you skew the triangle eg. to make a right-angle triangle, you don't change the areas. If you uniformly scale the triangle vertically by r (preserving the ratios between the areas), and then scale horizontally by 1/r, each region's area is scaled by the same factor.

[Genady]

16 minutes ago, md65536 said:

I can't imagine ever noticing that, but now I see why it works for other triangles with the same areas. If you skew the triangle eg. to make a right-angle triangle, you don't change the areas. If you uniformly scale the triangle vertically by r (preserving the ratios between the areas), and then scale horizontally by 1/r, each region's area is scaled by the same factor.

More rigorously,

Spoiler

the big triangle with the area 4+2 and the small triangle with the area 2 have the same base. Thus, the ratio of their areas equals the ratio of their heights. The same holds for the ratio of the triangles 3+x+y and x.

Now, these triangles have common heights, i.e., the triangles 4+2 and 3+x+y share the height, and the triangles 2 and x do too. Hense the above ratios of their areas are equal.